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Show that if C is the boundary of the triangle with vertices at the points $0,3i$ and $-4$ oriented in the counterclockwise direction, then $$\bigg|\displaystyle \int_C(e^z-\overline{z})dz\bigg| \leq 60 $$ I know that the length of triangle is 12, how can i get upper bound in such way i get above inequality.plz help me.thanks in advance.

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3 Answers 3

up vote 1 down vote accepted

With $\,z=x+iy\,$:

$$|e^z-\overline z|^2=|e^xe^{iy}-(x-iy)|^2=|(e^x\cos y-x)+(e^x\sin y+y)i|^2=$$

$$=e^{2x}(\cos^2y+\sin^2y)-2xe^x\cos y+x^2+2ye^x\sin y+y^2=$$

$$=e^{2x}-2e^x(x\cos y-y\sin y)+x^2+y^2$$

But $$-4\leq x\leq 0\;\;,\;\;0\leq y\leq \frac{3}{4}x+3\leq 3\,\,\,,\,\,\text{so:}$$

$$|e^z-\overline z|^2\leq 1+2(0+3))+9= 16\Longrightarrow|e^z-\overline z|\leq 4$$

Thus:

$$\left|\int_C|e^z-\overline z|\,dz\right|\leq \max_C|e^z-\overline z|\cdot 12=48$$

Disclaimer: Check the inequality three lines above.

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in last step why you are writting two times modulus? –  Siddhant Trivedi Oct 27 '12 at 16:16
    
I don't understand: in the last step I wrote 12 times the modulus, not two times...? –  DonAntonio Oct 27 '12 at 17:56

Since $e^z$ is entire and $C$ is closed, $$ \int_Ce^z\,\mathrm{d}z=0 $$ Furthermore, the length of $C$ is $3+4+5=12$ and $|\bar{z}|\le4$ on $C$, therefore $$ \left|\int_C(e^z-\bar{z})\,\mathrm{d}z\right|=\left|\int_C\bar{z}\,\mathrm{d}z\right|\le\int_C|\bar{z}|\,\mathrm{d}s\le4\cdot12=48<60 $$

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Note that $\int\limits_C(e^z-z)dz=0,$ because $f(z)=e^z-z$ is an entire function (analytic in whole complex plane $\mathbb{C}.$) Therefore, integral can be simplified: $$I\overset{def}=\displaystyle \int\limits_C(e^z-\overline{z})dz= \int\limits_C(e^z-z+z -\overline{z})dz= \int\limits_C(e^z-z)dz+\int\limits_C(z -\overline{z})dz= \int\limits_C(z -\overline{z})dz=i \int\limits_C \Im(z -\overline{z})dz=2i \int\limits_C \Im (z) \ dz =2i \int\limits_{C_1\cup C_2\cup C_3} \Im (z) \ dz .$$ On $C_3 =\{z=x+iy: \quad -4 \leqslant x \leqslant 0, \, y=0 \}$ $$\int\limits_{C_3} \Im (z) \ dz=0 .$$ Hence $(z=x+iy)$, $$\left|I\right| \leqslant 2\left|\int\limits_{C_1\cup C_2} \Im(z) dz\right| \leqslant 2 \int\limits_{C_1\cup C_2} |y|\cdot|dz| \leqslant 2 \cdot \max\limits_{C_1\cup C_2}(|y|)\int\limits_{C_1\cup C_2} |dz|= \\ 2\cdot{3}\cdot{\operatorname{length}(C_1\cup C_2)}=2\cdot{3}\cdot (3+5)=48.$$

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